A391469 Number of words of length 3*n that can be formed with a three-letter alphabet when the number of letters of each type is == 1 (mod 3).

0, 6, 90, 2106, 58806, 1596510, 43053282, 1162202418, 31380882462, 847290203766, 22876797237930, 617673353237226, 16677181570526406, 450283907053258830, 12157665462543713202, 328256967363156018018, 8862938119558357917102, 239299329231464818199526

Offset

0,2

Comments

See comments in A391468.

Links

Pablo Serra, Table of n, a(n) for n = 0..100

Pablo Serra, On the number of words of N = 3*M letters with a three-letter alphabet, arXiv:2512.22362 [math.CO], 2025.

Index entries for linear recurrences with constant coefficients, signature (27,-27,729).

Formula

a(n) = Sum_{n1+n2+n3=n-1} trinomial(3 * n; 3 * n1 + 1,3 * n2 + 1, 3 * n3 + 1) for n>=1.

a(n) = 3^(3*n-2) - ((1 + (-1)^n) + (1 - (-1)^n)*i*sqrt(3))*i^n*3^((3*n-2)/2)/2 for n>=1, where i is the imaginary unit.

a(n) = 27*(a(n-1) - a(n-2) + 27*a(n-3)), for n >= 4.

G.f.: 6*x*(1 - 12*x - 27*x^2)/((1 - 27*x)*(1 + 27*x^2)).

a(n) + A391468(n) + A391470(n) + 2*A013733(n-1) = 3^(3*n) for n>=1.

E.g.f.: (2 + exp(27*x) - 3*cos(3*sqrt(3)*x) + 3*sqrt(3)*sin(3*sqrt(3)*x))/9. - Stefano Spezia, Dec 30 2025

Example

For n=1, the a(1) = 6 words of length 3*n = 3 are ABC, CAB, BAC, CBA, BAC, CAB.

Mathematica

LinearRecurrence[{27, -27, 729}, {0, 6, 90, 2106}, 20] (* Paolo Xausa, Jan 17 2026 *)

Crossrefs

Cf. A391468, A391470, A013733.

Sequence in context: A379255 A004996 A001499 * A147630 A221097 A177584

Adjacent sequences: A391466 A391467 A391468 * A391470 A391471 A391472

Keyword

nonn,easy

Author

Pablo Serra, Dec 29 2025

Status

approved